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This is a follow up question to another question I made earlier.

I put some of this information into my own notes just so I can get my head around this.

So just to clarify. Does the Woodward Hoffman rule given below only apply to ground state reactions, thus the rule cannot be applied to the above reaction scheme because one of the components is in an excited state?

A ground-state pericyclic change is symmetry-allowed when the total number of (4n+2)s and (4n)a components is odd.

EDIT: (1) Correct labeling of the suprafacial components, (2) Removal of orbital phases, (3) Stereochemistry situated at the correct stereocentres. 14+2 stereoedit

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    $\begingroup$ Your diagram reveals some misunderstandings. (1) As you've drawn it both components there are suprafacial, not antarafacial. (2) You shouldn't be drawing orbital phases in a Woodward-Hoffmann analysis. Orbital phases are important for a Fukui FMO analysis, but not for W-H. (3) Stereochemistry in the product is missing, and the stereochemistry is important, because it tells you whether the reaction is s+a or s+s. If it's s+s (photochemically allowed), then the protons in the product must be cis to one another, like this: i.sstatic.net/gkOSu.png $\endgroup$ May 7, 2017 at 10:38
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    $\begingroup$ Thank you for your help (othocresol & NotBaran). I've been having a difficult time understanding pericyclic reactions. (1) For some reason, I was defining the supra and antara based on the orbital adjacent to each other rather than between components (2). I figured out how to draw a non-shaded orbital in Chemdraw. (3) That was silly of me not to show the stereochemistry at the stereocentres. $\endgroup$
    – Trufflehog
    May 7, 2017 at 12:51
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    $\begingroup$ The supra/antara is a common mistake when everybody starts out. I can confirm that it has confused me many times ;) $\endgroup$ May 7, 2017 at 13:35

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For reactions in the excited state, a modified statement of the Woodward Hoffman rule applies:

A pericyclic change in the first excited state is symmetry allowed when the total number of (4q+2)s and (4r)a components is even.

The only difference is that we're now looking for the total to be even rather than odd.

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